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Does Oswald Efficiency make a significant difference on RC-aircraft?
As we know, Oswald Efficiency ($\epsilon$) is caused by change in downwash over the span of the wing, thus causing a change in effective angle of attack over the wing and therefore induced drag variation across the wing.
Formula: $C_{di}=\frac{C_l^2}{\pi \epsilon R}$
For this reason, old planes and some hobby planes use elliptical wings (best seen by the Spitfire), so that smaller chord wingtips generate the same downwash as high-chords mid-wing -- keeping constant downwash and effective angle of attack throughout the plane.

I'm curious about how varying different factors has an effect on Oswald Efficiency, for example for an RC-Airplane, lets say roughly 1.5m span, 25m/s max velocity.
My intuition:
On one hand, having such little space between the wing tips and fuselage means less distance for vortices formed by downwash at mid-wing to form, which should allow downwash to be somewhat even. Further, the air at the wing tip would also "drag along" air closer in the center, kind of balancing out any differences in downwash and increasing Oswald Efficiency.
On the other hand, the pressure gradient above/below the wing  grows by velocity squared, and would be much lower for a 25 m/s plane. I'm not sure the force/pressure difference would be strong enough to induce strong vortices mid wing. Thus, there would only be small vortices at the wingtips, resulting in a very large downwash gradient, thus lower Oswald Efficiency.

Yes and no.  All the physics still hold.  The absolute distance from wingtip to fuselage doesn't matter -- flow structures scale down to match the aircraft.
However, most R/C aircraft are not flown like full scale aircraft.  That is -- most RC aircraft are not flown in trimmed equilibrium flight for long straight segments.
Instead, most RC aircraft are flown at a high throttle setting and the aircraft frequently are turning as we generally enjoy poking holes in the sky.
Maximizing fuel economy, range (or even endurance) is usually not the point of an RC aircraft, so achieving maximum flight efficiency is usually not a problem.
In addition, most RC aircraft are substantially overpowered as compared to their full-scale counterparts.  So places where low induced drag might be noticed are generally over-run with sheer might.
Where I have most noticed high induced drag (from Oswald type effects, but also low-span (or low aspect ratio)) -- is with an aircraft that barely has enough power that also has to make relatively sharp turns.  Such an under-powered aircraft will lose substantial altitude in a turn (when induced drag is increased).  If the aircraft is flying a racetrack pattern, it will lose altitude in the turn, struggle to climb down the straight, only to lose the altitude again in the turn.

Very good answer!

Thanks for you answer. You mention that most RC aircraft are not flown like full scale aircraft, and while I agree with you, there are plenty of applications for RC aircraft such as delivery drones as well as research/competitions where maximizing range/endurance does matter, so you're not really answering the question... Further, on RC planes, typically induced drag is a far bigger factor in max speed than pressure drag, and actually is asymptotic because pitching the plane angles your thrust, thus it's not something that can be "over-run with sheer might" as you say.

@Ankit, I stand by my statements.  I don't consider an aircraft (drone) that is controlled by an autopilot to be an R/C aircraft anymore.  The important difference is that an autopilot may actually attempt to throttle back and trim the aircraft to fly in equilibrium flight.  A true visually tracked and thumb-stick controlled radio controlled aircraft, the pilot has no airspeed, altimeter, or rate of climb instrumentation.

The pilot can not trim the aircraft to a target airspeed (best endurance, best range, etc) because they don't have the information.  Also, when flying within visual range as R/C aircraft do, you never fly in a straight line before needing to turn.  I have advised many student groups who have built RC aircraft for competition, I know how those usually work out.  Induced drag is never a big factor in max speed -- by definition.  Calculate the thrust-to-weight or power-to-weight of most R/C aircraft and compare them to their general aviation equivalents, you will find RC aircraft are overpowered.

I understand that you might not have felt that I answered your question.  Your intuition for what happens to aerodynamics when an aircraft is scaled down is wrong.  One of the fundamental ideas of aerodynamics is that you can scale a body up or down and the flow structures remain essentially changed.  This is why we can test large aircraft with smaller models in wind tunnels.  You need to match Reynolds number (Mach number is low enough here to ignore).  So any difference you see between model and full scale should be a viscous drag / boundary layer thickness / Reynolds number related.

@RobMcDonald Yep, exactly, Reynolds number scaling is a good measure for these kinds of things. I thought the Spitfire and hobby/passenger aircraft had an Re several order of magnitude greater than RC-scale planes, and thus a significant difference in effect. But since you know so much on the topic, why don't you enlighten me :)

Yes, their Re is much larger, but all the effects you described -- and the primary concept behind elliptical wings and Oswald efficiency -- are all inviscid concepts.  I.e. they are independent of Reynolds number.  Re will mostly effect the stall behavior of the airfoils (at what lift coefficient and whether it is abrupt or gentle) and the viscous drag.  Many scale RC aircraft use different airfoils from their full-size counterparts to improve stall behavior at low Re.


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Stall on low airspeed with high power on right-turning propeller
Q: The pilot of an aeroplane with a single engine and right-turning propeller (clockwise from the pilot’s viewpoint) decides to go around. The pilot raises the nose too much as the power increases and stalls the aircraft. Most probably, the aircraft will…
A: Enter an incipient spin to the left due to roll and yaw caused by the propeller.
I understand why aircraft will roll and yaw to the left; it is due to propeller effects, but how rolling and yawing to the left helps developing a spin to the left? One thing that comes to my mind is when yawing left, airflow on the right wing is higher then on the left wing so left wing is more stalled then right and this couses aircraft to spin left. Am I right?

Yes. How do you enter a spin? By bringing the plane to a stall and stomping rudder as the stall is starting to break. Left rudder, right wing speeds up slightly, so AOA is reduced slightly, left wing vice versa, so as the stall breaks, left wing is more deeply stalled than the right, and the spin to the left starts as the not quite fully stalled right wing wants to drive (glide) forward.
Here in Canada, spins are taught in flight training, so the effects of yaw at spin entry becomes pretty obvious to the student in short order.

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In the United States, is there any way I can fly without a medical certificate?
I live in the United States. I haven't been able to get a medical certificate yet, but I'm interested in flying airplanes (or maybe other kinds of aircraft). Is there any way I can fly without having a medical certificate? What are my options here?

Yes, BasicMed
https://www.faa.gov/licenses_certificates/airmen_certification/basic_med/

There are lots of options for flying without a medical certificate in the United States. Most of the regulations about medical certification are in 14 CFR Part 61, Section 23.
Note that even if an operation does not require a medical certificate, it's still illegal to fly if you have a medical condition which would make it unsafe for you to fly.
Your options for flying without a medical certificate are:
Fly with an instructor, or someone else as the pilot in command (FAR 61.23(a))
You usually need a medical certificate to act as the pilot in command, but you don't need a medical certificate to merely handle the controls!
This means that you can take flying lessons without a medical certificate, as long as you have an instructor with you. The instructor will act as pilot in command, but you will get to do some (or maybe even all) of the flying.
This also means that if you have a friend who's a pilot (with a pilot's license and a medical certificate), and you and her are in an airplane together, she can let you do all of the flying! She will need to act as pilot in command, meaning that she's responsible for ensuring that everything is done safely, and for taking control of the aircraft if need be. But she can have you do the flying.
In case you're working towards a pilot's license, flying lessons received from a certified flight instructor can be logged as pilot time. If you're flying with a friend who's not an instructor, you cannot log that time.
Fly a glider or a balloon (FAR 61.23(b))
You don't need a medical certificate at all in order to fly a glider or balloon. Even without a medical certificate, you can still:

Get a solo endorsement from an instructor and fly by yourself.
Get a pilot's license and fly with passengers.
Get a commercial pilot's license and fly for compensation.
Get a flight instructor certificate and give flying lessons. (In a balloon, you only need a commercial pilot's license to give flying lessons; there is no flight instructor certificate for balloons.)

If your goal is to get an airplane pilot's license, pilot time in balloons and gliders can be counted towards that license! To get a private pilot's license for airplanes, you need 40 hours of pilot time; some, but not all, of that time is required to be done in airplanes.
Fly a light-sport aircraft (FAR 61.23(c)(1) and (2))
If you have an aircraft whose maximum takeoff weight is 1,320 pounds or less, it's probably a light-sport aircraft. If you fly a light-sport aircraft under the sport flying rules, then you can satisfy the medical requirement using a driver's license instead of a medical certificate. If your driver's license has any limitations, then those limitations also apply to your flying.
However, there's a caveat. In order to fly a light-sport aircraft using a driver's license instead of a medical certificate, you must satisfy either of the following conditions:

You have never applied for a medical certificate.
After the most recent time that you applied for a medical certificate, you were issued the certificate, and that certificate has never been suspended and revoked.

This means that if you're a licensed sport pilot, and you don't have a medical certificate, and you apply for a medical certificate, the privileges of your sport pilot's license go away until the medical certificate is issued!
Time in light-sport aircraft can be logged as pilot time.
Fly under BasicMed (FAR 61.23(c)(1) and (3), and 61.113(i))
If you've had a medical certificate in the past, but it's expired, you can still fly if you meet all the requirements for BasicMed. The main requirements are:

You must have a driver's license.
If your driver's license has any limitations, then you must also obey those limitations while flying.
You must have held a medical certificate at some point after July 14, 2006.
Your doctor must do a medical examination and sign off.
Your most recently issued medical certificate must not ever have been suspended or revoked.
Your most recent application for a medical certificate must not have been denied.
You must comply with the operating limitations in FAR 61.113(i).

Fly an ultralight aircraft (FAR 103.7(b))
If you're brave enough to fly an ultralight aircraft, good news! You don't need a medical certificate (or even a pilot's license) to fly a Part 103 ultralight aircraft.
Time in ultralight aircraft cannot be logged as pilot time.

Regarding the first section about flying with an instructor, I agree as long as you consider them orientation flights.  However, per 61.23(a)(3)(i), if you want to log your flight time with a CFI you would need a 3rd class medical and student pilot certificate.  As far as taking the controls with any pilot other than a CFI, I know this happens all the time but I would challenge you to find it specifically permitted in the FARs. And wait, did you both ask and answer the question @Tanner Swett?!

For the light sport aircraft case, it's important to mention the condition that you "Not know or have reason to know of any medical condition that would make that person unable to operate a light-sport aircraft in a safe manner." If the reason you don't have a medical certificate yet involves some medical condition (as opposed to "I haven't made an appointment"), this route may not be legal and/or safe.

@MichaelHall Re logging flight time without a medical: I don't think that's correct. 61.23(a)(3)(i) says that you need a medical certificate "When exercising the privileges of a [...] student pilot certificate", but receiving flight training is not one of "the privileges of a student pilot certificate". 61.51(h)(1) says that "a person" (not "a student pilot") may log training time. Re letting the passenger take the controls: as far as I know, it's permitted because no regulation prohibits it. See https://aviation.stackexchange.com/questions/3039/.

@ZachLipton I think that point is covered by my note that "it's still illegal to fly if you have a medical condition which would make it unsafe for you to fly". (61.53(b) and (c) places the same restriction on all non-Part 103 operations; I don't know why the restriction is restated for light-sport aircraft in 61.23(c)(2)(iv).)

@Tanner Swett - Interesting...  I will concede your point because the FARs don't make a link between "exercising the privileges of a student pilot certificate" and an uncertified "person" logging training time with a CFI.  Personally, however, I don't see the distinction.  Also, I too subscribe to the philosophy that if something isn't specifically prohibited then by default it is permitted.  You seem to have a very solid grasp of the FARs, so again, why did you ask the question?  Are you fishing for up-votes to boost your reputation number?!

@MichaelHall I posted this question and answer because I thought other people were likely to find it useful and interesting.

I respect your answer in terms of stimulating discussion, but you are being disingenuous.

@MichaelHall If you think I'm being disingenuous, please create a meta post, or send a chat room message, so that we can discuss this.

I don't know what a meta post is, and I couldn't find a link to private message you.  It isn't a matter of my personal opinion, your first post meets the very definition of disingenuous.  There's no malice in it, and you didn't try to even conceal your identity in the response, but that doesn't mean that the original question was not disingenuous since you clearly knew the answer already.  Again, you've provided some very useful information, but if your goal is to provide instruction in the FARs there are other methods...

@MichaelHall Here's the chat room, for general discussion; I check it pretty frequently: https://chat.stackexchange.com/rooms/12036/the-hangar Here's the meta site, which is for posting questions about this website (such as "is it appropriate to post a question you already know the answer to?"): https://aviation.meta.stackexchange.com/

Thanks for the links. I have no beef with you over this, and am letting go. I just thought it was a little odd when I realized I was debating with the same person that asked the original question!  :)  Cheers,

@MichaelHall Not a problem, thanks for the feedback.

Why is balloon time loggable as pilot time, but not ultralight time?  That seems backwards, given that an ultralight is more similar to a typical airplane than a balloon is.

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Does the pressure at the static ports drop as the aircraft's speed increases?
This answer by Tyler Durden explains what static ports are and how they work:

The "static ports" allow air into the instrument to provide the input for the air pressure on the side of the aircraft. They are called "static" ports because in general they take in air at the ambient pressure which does not change very much [...]

Currently, the answer has a comment by Zeiss Ikon which states:

In fact, the pressure at the static ports will drop as the aircraft's speed increases (Bernoulli principle) [...]

Is this true? If I'm flying at a constant altitude, will the pressure at the static ports decrease as I speed up and increase as I slow down?
If so, does the altimeter have a mechanism for compensating for this?

Something I keep coming back to -- one great way to learn something is make a comment based on what you know, to find out what you don't know.  Thanks for asking this!

The comment:

In fact, the pressure at the static ports will drop as the aircraft's
  speed increases (Bernoulli principle)

is not exactly accurate. The aircraft is a complex aerodynamic shape and static pressure distribution across the fuselage is not linear. An example is provided by the following figure from NASA Technical Memorandum 104316:

[Source: NASA]
If you look at the graphical part of the above figure, you'll see areas where the pressure is above (>0) and below (<0) the reference pressure. As the airspeed increases, the tendency will be for the positive values to get bigger and the negative values to get lower (more negative). So along the line between points 2 and 3, and also between 4 and 5 (over the wing), the pressure will drop as speed increases. Between 1 and 2 (the nose), and between 3 and 4 (just in front of the wing), the pressures will rise with speed.
From the same report:

Zero static pressure error on the fuselage exists at locations 2
  through 5. One of these locations is chosen for the static port. To
  keep pneumatic lag small, the static port is normally located as near
  the airdata instruments as possible (or the other way around). (To
  determine this location precisely, several static ports are made in
  this area. The optimum location is then selected as a result of
  comparing the various ports with a reference source, such as a
  trailing cone.) This pressure distribution changes with flight
  condition, so a calibration over the flight envelope may still be
  necessary.
  ...
  Even with the selection of the best static port position, some
  pressure errors will remain, and these errors must be determined in
  flight. The difference between the locally measured static pressure
  and the ambient static pressure, which is dependent upon angle of
  attack, airspeed, and aircraft configuration, is called position
  error.

Based on the measured position error, a Static Source Error Correction (SSEC) table is developed. The SSEC is loaded into the air data system when it is installed in the aircraft so that it can make the necessary corrections.

This is good to know! The last paragraph you quote states that position error "is dependent upon angle of attack, airspeed, and aircraft configuration"—so it does, in fact, have a dependency on airspeed. Does it depend on airspeed in a somewhat predictable way (such as "when airspeed increases, measured static pressure decreases", as Zeiss's comment suggested), or is it mostly unpredictable?

Yes, the correlation between measured airspeed and the adjustment required to the static port reading is what goes into the Static Source Error Correction table.

Note that this requires the altimeter system to know the measured airspeed in order to compensate, so a stand alone altimeter will likely not adjust out any position error unless it has some way to get that from an airspeed source.

@TannerSwett  The variation isn't predictable by analytical means, hence the need for flight test. It is however consistent and repeatable which is why once it is measured and documented in the SSEC the system can produce an accurate altitude.

I'm still wondering one thing, though. Is there any kind of general, overall tendency for the static pressure to read low when airspeed is high? Or is there no such thing?

@TannerSwett  If you look at the graphical part the figure, you'll see areas where the pressure is above (>0) and below (<0) the reference pressure. As the airspeed increases, the tendency will be for the positive values to get bigger and the negative values to get lower (more negative). So along the line between points 2 and 3, and also between 4 and 5 (over the wing) the pressure will drop as speed increases. between 1 and 2 (the nose), and between 3 and 4 (just in front of the wing) the pressures will rise with speed.

For a basic shape, such as a sphere, the pressure on the side of the object would indeed decrease due to Bernoulli's principle. However, as Gerry's answer points out, a plane is much more complex aerodynamically. The graph in Gerry's answer shows how pressure decreases at some points, increases at others and is unchanged in some.
For indicated airspeed, a small error could be adjusted out easily enough. After all, the pressure difference would still be mostly linear in relation to air speed, so just calibrating the scale on the airspeed indicator would suffice.
But the static ports are also used for the altimeter. For this reason there is a strong incentive to find a location that minimizes the so called position error in the static pressure measurement. Otherwise when indicated airspeed changes, the indicated altitude could also change, which would be much more difficult to compensate for.
So in summary: no, for a well-designed plane, the speed-induced pressure change at static ports will be minimal.

There is a difference between the true static pressure and the measured static pressure, which varies with angle of attack, but it is generally quite small.  This is called Position Error.  The placement of the static port is at the spot where the Position Error is the smallest through the airplane's normal operating attitudes.  Position Error has to be small enough to keep the measured pressure altitude within 30 feet of true pressure altitude, and an error of not more than 3% or +/- 5kt on airspeed.

I don't see how this answers the question @JohnK, the OP is asking about pressure dropping due to airspeed.


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Was the 1992 Gimli DHC-4 Caribou accident ever investigated?
On August 27, 1992, 3 people were killed when a de Havilland Canada DHC-4 Caribou crashed immediately after takeoff at Gimli Industrial Park Airport.
Wikipedia describes the accident as follows:

Three people were killed on August 27, 1992 when a NewCal Aviation turbine-modified de Havilland Canada DHC-4 Caribou they were aboard crashed on the airfield during climb-out after a short take-off from the airport. The aircraft nosed sharply up, arced right and nosed into the ground. […] The cause was listed as failure to deactivate the plane's gust-lock control in the cockpit, […] [T]he Transportation Safety Board of Canada has no record of this incident.

The entire flight was captured on a video which has circulated on the Internet.
Was this accident ever officially investigated? If not, why not?

It may have been (and probably was) investigated by the Canadian TSB, but the report is probably not available online as it was during the time that they were "digitizing" reports.

The online report database is supposed to cover 1991 and later incidents, but this particular report is not there.  The mystery is how they were able to take off with the gust lock engaged, since on the Bou and the Buff (same flight deck) the gust lock T handle latches engaged when pulled aft and it blocks the throttles from being moved out of idle.

It was, but from what I can tell the report has not been digitized yet.
The gust locks were left installed. The aircraft was being operated in the restricted category as they were retrofitting with the PT-6’s.
Here’s an analysis from James Donnelly, Bombardier Aerospace. Starts on page 7.
http://www.asasi.org/papers/2001/Four%20Unrelated%20Accidents.pdf
Essentially the rigging was done incorrectly. The gust locks did reduce engine thrust, but didn’t kill it completely. Takeoff roll was 20% longer than calculated.

The above answer has errors:

The gust locks were not "left installed"; they were left activated. The DHC-4 has internal gust locks, not external. Said gust lock interacts with the throttle levers in that the gust lock, when activated, normally prevents any advance of throttle levers but this is only valid on a non-modified, piston-powered DHC-4; in this particular DHC-4, converted to PWC PT-6A turboprop engines, the throttle levers--called power levers for turboprops--now, by modification, did not interlock with the gust lock control and now worked fully independently of said gust lock, whether it was activated or not.
"The rigging"? Please elaborate: that of the throttle/power levers or gust lock? It is assumed the throttle levers were replaced by new power levers which were not made to interact with the activation of the gust lock as the previous throttle levers had been, by-design.

It is uncertain whether or not the crew of this aircraft's fatal flight was familiar with this particular aircraft and its modification, or with the DHC-4 generally; what is known is that the gust locks remained activated, the power levers were advanced, the aircraft rolled a remarkably short distance, rotated, pitched up dramatically, climbed over 100 feet, achieved near-vertical pitch, rolled right 90-degrees at apex, continued on, gained speed and nosed into the ground at full power, impacting at a steep angle. All aboard died on impact and the aircraft caught fire shortly after. It should be noted the aircraft did not explode but merely caught fire.
This answer may be wordy but sometimes extra words are necessary to clearly convey fact. Economy of words for the sake of brevity is poor policy.

Do you have any references for this information?

You shouldn't say "the above answer", even if you firmly believe that your answer will never be upvoted to have more votes than the other answer, because answers can be sorted in several different ways.

The accident was investigated and a report was published the following year, in 1993. The accident report apparently was not available online at the time that the question was posted, but it is available now: investigation number A92C0154 at tsb.gc.ca.
The accident report concluded that

The gust lock system was not fully disengaged prior to flight and one or more of the gust locking pins became re-engaged for undetermined reasons after lift-off. It is unlikely that a control check had been completed prior to take-off and, once airborne, the crew were unable to disengage the gust lock mechanism before losing control of the aircraft.


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Do airline pilots use speed stability to cope with unreliable airspeed indications? If not, why not?
There have been a couple of airline accidents in which unreliable airspeed indications (or the fear thereof) have played a role:

Birgenair Flight 301 in 1996, where one of the pitot tubes was blocked, causing unreliable airspeed indications. At one point, the captain gave full throttle at low airspeed and high angle of attack, causing the left engine to flame out, and producing a spin.
Aeroperú Flight 603 in 1996, where the static ports were blocked, causing unreliable airspeed and altitude indications. The altimeter read high, and the plane crashed into the ocean.
Air France Flight 447 in 2009, where the pitot tubes iced over, causing unreliable airspeed indications, for about one minute. The aircraft entered a stall and the pilots failed to recover from it. Presumably, the pilots were uncertain about whether or not the airspeed indications were reliable, which contributed to their failure to recover.

Now, correct me if I'm wrong, but if your airspeed indicator fails in, say, a Cessna 172, you can use the aircraft's speed stability to maintain a safe airspeed. Do something like:

Move the trim wheel to the takeoff position.
Keep the nose from rising or falling suddenly (unless you're near a stall), but allow it to rise and fall slowly.
If the pitch stays constant with no elevator input, you're flying at a reasonable speed. If it tends to rise, you're flying faster than that, and if it tends to fall, you're flying slower than that.

Once you get the nose to stay still, you don't have to touch the elevator much unless you change power or configuration.
Yet, I've never heard of airline pilots doing anything like this. Do they do this sometimes, and if not, is there a reason why they don't?
(For that matter, is there a reason why my suggested procedure isn't a good idea in GA aircraft either?)

Although adding power helps, the key is to reduce AOA.  Vertical sink adds to AOA at a given pitch relative to horizon.  H stab will help nose down in a sink.  Good design helps here.  I would trim elevator only after sink is arrested at a higher power setting, then set rpm and trim for level flight.

Why do you suggest takeoff setting instead of cruise?

@TomMcW Just because the takeoff position is clearly marked.

@Tanner Swett, definitely on the right track with pitch setting.  Power input will vary from plane to plane.  GA props and higher performance, probably best left alone until stall is broken.  I have heard of tri-jets just powering out of it, low slung jets may actually increase pitch up with power.

Longitudinal stability is certainly useful part of flying with unreliable airspeed, but far from sufficient. The problem is that you can only trim for current speed, but you need to trim for cruise speed first, and then for landing speed.
There are many factors that affect the trim of the aircraft. The most important are: weight and balance, engine power and the position of the trim tabs and/or stabilizer.
The weight and balance matters a lot as evidenced by this incident. That means take-off trim is mostly useless—it is just some middle value so that the actual trim needed for take-off is not outside the authority of the elevator, but the actual deflection needed will be different from flight to flight.
So instead the unreliable airspeed procedure calls to fly level with specific pitch attitude—which results in reasonable cruise speed—and then reduce power until specific rate of descent is achieved while maintaining the pitch attitude—which results in reasonable landing speed.
Also in case of Airbus, while the angle of attack protection is lost under unreliable airspeed condition, the auto-trim function remains operational, so the aircraft still behaves as neutrally stable. This makes the procedure slightly easier, as when power is adjusted, it will keep flying straight and raise or lower nose on its own, but also slightly more risky, since there won't be pitch-down tendency before stall.
After the Air France flight 447 accident, Airbus introduced ability to show angle of attack instead of airspeed in case airspeed is unreliable, which makes the procedure much easier.
That is unreliable airspeed. Blocked static ports, as in the case of Aeroperú flight 603, are much, much worse, because they render all of airspeed, altitude and vertical speed inoperative, and vertical speed is much, much more critical. The pitch-power tables should allow setting power that will result in mostly level flight given weight and approximate altitude, but it is a lot harder to maintain safely.

I don't believe the procedure you've described is formally recommended. Instead there is the unreliable airspeed procedure. The exact parameters vary among aircraft types but the basic principle is that by setting a specific pitch and power combination, you're guaranteed a certain airspeed/performance.
These procedures have existed for many years but perhaps the biggest change to come from the AF447 crash is an increased emphasis on this in recurrent training.
See: What is the correct crew procedure for an unreliable airspeed alert?

Well, to be clear, the course of action I'm asking about is to set the trim to a known good setting, and then use the elevator for the sole purpose of stabilizing the pitch, with the goal of taking your hands off the stick and letting the aircraft maintain whatever pitch it happens to maintain. In other words, let the trim determine the pitch. The procedure you linked to is something different: namely, to intentionally maintain a specific pitch, and then use trim as needed. So that means the answer to my question is "no, airline pilots don't do that", right?

By the way, you write that "by setting a specific pitch and power combination, you're guaranteed a certain airapeed/performance". But—and I might be seriously misunderstanding something here—doesn't AF447 show that that's not the case? AF447 was flying "on the back of the power curve", so that even though the pilots flew at more or less the correct pitch and power, they got the wrong performance. If I remember right, at one point they had the power at TOGA and the nose 10 degrees below the horizon, and even this failed to increase their airspeed enough to break the stall.

@Tanner Swett - OK, I have updated the answer. I'll let someone else comment on whether it is feasible in an airliner. But as for AF447, the crew needed to use the unreliable airspeed procedure as soon as the problems started - in fact the best course of action would have been to do nothing at all. But the first officer pulled his sidestick back full and kept it there until pretty much the impact. The guaranteed performance I mentioned won't work when the stall was that severe, at less than 60 KIAS that aircraft was a brick and positive action needed to be taken.

The stick being pulled all the way back is exactly why it was stalled.  Orientation to the horizon really has nothing to do with it.  You can stall anywhere in a loop if AOA is too high.  Angle of AOA determines stall, which causes loss of lift.  If a plane is moving forwards at 400 knots but sinking at 300 knots with full aft stick it will be stalled.  Relaxing back pressure is key.  Crews must be trained and kept current.  The pitch setting was not correct.

@RobertDiGiovanni: If your aircraft is moving forwards at 400 knots and sinking at 300 knots with full aft stick, you're travelling at 500 knots at a very high attack angle, and probably breaking apart in midair.

@Sean that is true!  This is why stall recovery must be unloading the elevator (and pushing forward if needed) rather than trying to "power out".  Never exceed speed is the limit, better to point the nose further down with no power and let gravity help get a lower AOA for the wing. But I did change the numbers in later writings to 250 knots forward and 125 knots down! Thanks.

From a basic airmanship perspective, if you are plodding along minding your own business in an airliner and all three airspeed tapes go wonky without any other changes to sound or movement, thrust, trim, etc, your basic instinct should be to do nothing and evaluate other instruments, mainly pitch and altitude, and steady as she goes.  If you've been flying the same plane a long time so that the general numbers are familiar, you should be able to achieve various speeds because you can recall the pitch attitudes and thrust settings for various configurations.  For example, I haven't flown an CRJ200 in quite a while but I still remember that fan RPM at Flaps 45 at Vref on a 3 degree glideslope is around 62% at MLW and around 57% at light weight and the nose attitude is about 2-3 deg down (the 200 and Challenger bizjets are weird that way). Same thing in other conditions; you know the plane well enough that you know what thrust setting and pitch gives you more or less the speed you want - good enough to keep you below Vmo and above stall at any rate.  There was a CRJ crew at Atlantic Southeast Airlines in the late 00s that had a triple airspeed loss incident and they simply used really good familiarity with the type and good basic airmanship and handled it no problem.
That kind of knowledge wasn't sufficient in those accident cases as crews became mentally overwhelmed and slipped into panic mode (nobody really knows if they will or won't until it happens- good training just helps tip the probability of a good result in your favour), and in response the OEMs have all created pitch/power tables that go into the Quick Reference Handbook as part of an Unrealiable Airspeed QRH procedure.  Basically, tables, derived from flight test and analysis, that say for Pitch of X degrees, N1 setting of Y, with so and so flap setting and at so and so weight range, you get a speed of more or less Z.  The data is good enough to get you down safely.
On GA airplanes it's very useful to understand that the trim wheel is in fact your speed dial, like a cruise control setting on a car you might say.  The speed the airplane wants to seek is based on where the trim is set, modified slightly by power setting when the engine is in front and is blowing on the tail.  That is, adding power reduces trim speed on a tractor aircraft and reducing power increased it, by a few kts.
So if you are in a 172 and a wasp in the pitot kills your airspeed, as happened to me once, and you know roughly what power setting you typically use on final, and roughly what pitch attitude to use, and set the trim so it holds that, and the trim indicator is roughly where you expect it, you will be fine.  If you are really too slow it should become fairly obvious as things become quiet and mushy.  You aren't going to just spontaneously stall and spin if you are letting the plane do its thing at a reasonable trimmed speed.  The more common problem in my experience when GA pilots lose their airspeed is they overcompensate at the other extreme and end up landing at 100 mph.

Capt. Reynolds suggested data charts be gathered using engine RPM, and, as you suggested, trim condition. Charles Lindbergh stressed "being as one" with your aircraft.
The better you know your machine, the less reliant the pilot is on one instrument.
In this day and age we also have GPS tracking for speed.  Backups can be critically important, as is the common sense to be aware and avoid icing conditions if at all possible.
Ice buildup would not only block the pitot tubes (if heating was inadequate), but also lower the lifting ability of the wings, raising stall speed.  Additionally, icing on props will reduce their efficiency, resulting in less thrust at a given power setting.
You may wish to set trim a little closer to cruise and monitor the altimeter and RPM.
Then get on the radio and head for a large airport with radar tracking and a long runway.
They could help with the approach.
Lacking this, a long straight stretch of superhighway or large field would do for a landing
site (dry lakebeds too).

This does not appear to answer the question.

Paragraphs 5 and 6 are my suggestions, along with input from others.


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Speed change in IFR flight in class C or D airspace
Can I change speed while flying IFR in class C or D airspace without having a clearance from ATC?
I tried to find it in some documents on EASA website but I did not find anything.

Can you fly IFR in EASA controlled airspace without a clearance?

Im asking about Poland for example, but i would like to know how it looks like in different countries

Yes, as long as ATC did not assign you any speed restrictions and so long as you comply with any applicable chart-listed speed restrictions and with things like 14 CFR Section 91.117, obviously.
See AIM 4-4-12 'Speed Adjustments'.

Thanks for helping, what about european airspace? AIM is an FAA document.

I won't pretend to know the answer vis-a-vis European airspace - hopefully somebody else will come along who does.


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Why if you tune localizer frequency the glide slope is tuned too?
if tuning the ILS you tune localizer freq which is between 108 MHz to 111.975 MHz. why the glide slope freq which operates in the 328.6–335.4 MHz band is tuned simultaneously?

Are you asking how this works from a technical point of view or why it is done this way?

This question might also be of interest

A glideslope signal is useless without a localizer signal (i.e. if the localizer is out of service, the entire ILS is out of service). There is never a case when a pilot would be expected or required to tune a glideslope frequency unless they were tuning a localizer frequency at the same time. To make things easier, and to reduce the possibility for user error (the pilot tuning the wrong GS frequency for the desired LOC), the two frequencies are used in known paired sets and the airborne equipment will tune the correct GS frequency associated with the LOC frequency.
This is also how DME works; DME stations operate in the 960–1215 MHz band, but the pilot simply tunes the desired paired VOR frequency without having to enter the DME frequency as well. However in this case there are standalone DME stations not associated with VORs. These stations are still referred to in aeronautical publications by the associated VOR frequency.

There are also DMEs associated with ILS that are paired with the LOC frequency. SO tuning the LOC will also tune GS and DME.


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Can I land in ILS CAT I when no RVR info is available and VIS is 700m? According to easa
Category I (CAT I): a DH not lower than 60 m (200 ft) and with either a visibility not less than 800 m or a RVR not less than 550 m

I was wondering if it is legal to convert VIS to RVR and land when calculated RVR is above 550m

If the visibility was 800 meters or above when you entered the final approach segment or 1,000 feet and then dropped to 700 meters, you could still land if you have the required visual reference when you reach the DH of 200 feet.
If the visibility is 700 meters before the final approach segment or 1,000 feet, the approach ban would apply.
See "approach ban" and "required visual reference" EASA (CAT.OP.MPA.305)
For ease of reference

(a) The commander or the pilot to whom conduct of the flight has been
delegated may commence an instrument approach regardless of the
reported RVR/VIS.
(b) If the reported RVR/VIS is less than the applicable minimum the
approach shall not be continued:
(1)  below 1000 ft above the aerodrome; or
(2)  into the final approach segment in the case where the DA/H or
MDA/H is more than 1000 ft above the aerodrome.
(c) Where the RVR is not available, RVR values may be derived by
converting the reported visibility.
(d) If, after passing 1000 ft above the aerodrome, the reported
RVR/VIS falls below the applicable minimum, the approach may be
continued to DA/H or MDA/H.
(e) The approach may be continued below DA/H or MDA/H and the landing
may be completed provided that the visual reference adequate for the
type of approach operation and for the intended runway is established
at the DA/H or MDA/H and is maintained.
(f) The touchdown zone RVR shall always be controlling. If reported
and relevant, the midpoint and stopend RVR shall also be controlling.
The minimum RVR value for the midpoint shall be 125 m or the RVR
required for the touchdown zone if less, and 75 m for the stopend. For
aircraft equipped with a rollout guidance or control system, the
minimum RVR value for the midpoint shall be 75 m.

You may convert the visibilty to RVR according to the table, see (c) above.

I was wondering if it is legal to convert VIS to RVR and land when
calculated RVR is above 550m


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How do AeroVironment's Switchblades steer their flight?
I've recently been to a defense expo and saw a previewed Switchblade 300 at the AeroVironment booth. It got me questioning how the Swtichblade, and similar fold-out drones steer their flight. The Switchblade lacks ailerons, elevators, or rudders. So how do they steer? This question has been asked before, but I feel the answers were lacking. Here are pictures I took of the Switchblade 300.
Thank You

Ah looking at this, I realize that they no longer have ailerons/elevons... I will have to update my answer over at the other thread. I hope I can use your photos for updating my answer. My best guess is that they now tilt the entire wing for roll and pitch control.

After some looking around, I am convinced that this is a presentation model, which does not feature the ailerons which are normally used for flight controls. See this video of the Block 20 model with ailerons displayed https://www.youtube.com/watch?v=4pTMM9ZUpTQ.


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What are the "Aft Bobweights" for on a full power hydraulic aircraft?
I like military aircraft, and recently I've been researching the feel system for the control stick.
It seems that there is something called bob weights to improve the feel of an aircraft. It's a logical system to provide resistance to the stick according to vertical acceleration.
My question is that some hydraulically controlled aircraft have two bob weights.
Why do I need two bob weights? Looking at the diagram, it appears that the rear bob weight pulls the stick toward you.
In other words, does it mean that the mass balance is achieved by the front bob weight and the aft bob weight?

The bobweights provide two functions:

A artificial stick force gradient that rises with increasing G.
Inertial damping to the control circuit by adding mass to the linkage to dampen out sudden or jerky pilot inputs independent from G load. With hydraulic controls, the only feedback the pilot feels is whatever resistance is in the linkage and any springs. You could say the bobweights add "heft" to the control system that would otherwise be too easy to move. Viscous dampers are often included to further fine tune the feedback of the system. The objective is to try to replicate the feel of a normal manually operated control surface where the mass of the surface itself influences control feedback.

To achieve both objectives at the same time, you use two bobweights that oppose each other. The two weights combined provide the desired intertial effects the designer is looking for. For stick force-per-G, you make the nose-down bobweight heavier, or with a longer arm, than the nose-up one, giving you a net nose-down bias that provides the required stick force per G.

Wouldn't the mass of the pushrods give enough inertial damping? It seems silly to make them as light as possible and then add masses for feel. My version of a "bobweight" is the upgoing pushrod into a T-tail - you only need to make sure that the gearing works the right way.

Using dedicated weights allows you to find tune the effect you want by adjusting the mass and leave the control linkage alone. Modern transports with cable operated hydraulic controls do away with bobweights and use a pitch feel bungee spring (or hydraulic) devices with variable spring rates and damping. The Challenger bizjet just connected the feel units to the moveable horiz stab to vary the stick force by trim speed.  On the CRJ line, an outgrowth of Challenger, that was replaced with an electronic controller with linear input actuators.

Opposing bob weights apply forward and backward pulling force in response to vertical acceleration, so the back and forth movement of the stick becomes ``heavy''. understood.
And since the front bob weight is heavier, you feel a stronger force pulling you forward. understood.
It seems that stabilization by inertial mass can be achieved with just the front bob weight, but is a rear bob weight necessary for balance?
The rear bob weight is basically built into the tail of the aircraft. Is this also relevant?


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Why does aircraft weight influence position (installation) error in pitot-static systems?
I am a private pilot from Argentina studying to become a commercial pilot. In one of the books I was given to study, published by the Chilean air force, the there's a claim (in Spanish, page 36) that aircraft weight influences the installation error in a pitot-static system.
To me, this is counter-intuitive as I would think the weight of the aircraft should have no effect on the airflow around the pressure sensors outside the aircraft (pitot tube and static port). The only possible way I would think aircraft weight would have an effect on installation error, is if this weight implies a change in shape which in turn affects airflow, but I'm not sure this is the case.
I was able to track that statement back to where it was (probably) taken from, which is on this old website from FIU (partially funded by NASA), but no further, and I wasn't able to find any other source whatsoever mentioning weight as a factor.
If it's not mentioned anywhere else, it's probably not that important, but I want to know if I should adjust my mental model on the physics sorrounding how these measurements are made.

At a given speed, the lighter aircraft will have a lower Angle of Attack than the heavier one, so at an actual 200 knots (and level flight), the indicated airspeed will have more effects from the AoA (i.e. the pitot tube not aligned directly into the relative wind) with the heavier aircraft.

So it's not weight per se causing the measurement difference, but its influence on the AoA... so it's more of an indirect effect. That makes more sense, thanks.


text:
Help identify symbol of vintage biplane seaplane wing
I recently bought a "Vintage" biplane seaplane model from an Etsy shop. The images on the listing didn't include an unobstructed view of the top of the wings. When I received it, I found that it had symbols displayed on each end of the top wing. The symbol (attached) looks like an X with curved tips and a diamond in the center. My search so far has only returned similar listings for similar models, but I have been unable to locate the aircraft type or the origin of the symbol. My main concern is the origin of the symbol. Any help would be really appreciated!

It's highly likely that this is "art", and as such may be completely unrelated to any actual aircraft...

Explain why this has to do with aviation. It could be simple art.

@AndreasLauschke, it is aviation art. Asking whether it represents an actual airplane is not off topic. Merry Christmas!

The float-plane formerly known as Twitter ?

Strong pirate vibe, IMHO.

The symbol is unlikely to represent anything outside the builder's imagination, because so much of the rest of the airplane is unrealistic.

The outboard struts are canted inwards, not outwards.  (The upper wing is almost never "shorter" than the lower one, for many reasons.)

The inboard struts are between the wings, instead of being cabane struts.

The strutwork between floats and tail is overbuilt.

The floats are almost as big as the wings.

Tail floats are exceedingly rare.

The symbol might be hinting at the rounded cross used by German airplanes in 1914-15, without being too obviously German.  But only the builder knows for sure.
All that being said, the design is undeniably cute, rather like a puppy with big feet, so I'd say the builder has succeeded in what they were aiming for.

The float struts cannot take shear and - if that were a real plane - will fall off in an attempt to land on water.


text:
LPV, WAAS/EGNOS and GLS in 737 NG
As far as I understand, there are mainly 3 types of RNP approaches:

LNAV/VNAV - we have a set of fixes with altitudes and flying it
LPV WAAS/EGNOS - same as 1, but aircraft use special receiver equipment based on geostationary satellites, that's why it's so-called SBAS (sattelite-based)
LPV GLS - which is same as ILS, uses GPS + signals from ground equipment. It's so-called GBAS (ground-based)

Let's take EPRZ RNP Rwy 09.

Here we can see three options:

LPV CAT 1 - which I assume shall use EGNOS (V in LPV is vertical)
LNAV/VNAV - which I assume use only FIXes and altitudes
LNAV CDFA - which I assume uses LNAV and I need to descend somehow (V/S or glide path angle)

In terms of 737 NG, LNAV/VNAV and CDFA are straightforward.

Do I need to "tune" WAAS/EGNOS channel somehow? Referring to this LPV/LP approach (RNAV with WAAS): What is actually the "channel number"? it seems like no
If I don't need to tune EGNOS channel, what is the difference between LNAV/VNAV and LPV WAAS/EGNOS regarding the procedures? How do I know, if aircraft is ready for LPV with EGNOS/WAAS?
Approach is called LPV CAT 1, on the other hand, in EPGD RNP approach named just LPV. What is the CAT 1 here?
Can IAN be used (if equipped) instead of LPV, LNAV/VNAV and LNAV CDFA?
I assume something here is related to ANP/RNP, but chart has no RNP minimum values listed here. What are the general rules, I probably missed?

FCOM 737 says only about ILS/GLS (normal procedure) and supplementary SP 4.7 Instrument Approach - RNAV (RNP) AR, but in such case it's non-AR approach.
I'm not 737 pilot yet, using 737 in simulators only

Regarding your 5th question: What RNP level is required for an RNP approach if the chart does not specify a value?


text:
How does the horizontal component of lift when flying at a bank angle cause the aircraft to follow a circular path?
I understand that when the aircraft is put into some bank angle, the lift vector now acts at that angle instead of directly against gravity, so we can pitch the aircraft up to increase AoA to increase lift so that the vertical component of lift cancels out weight. Then we are just left with a horizontal component of lift acting perpendicular to the forwards direction of the aircraft. I don't understand why this doesn't just cause the aircraft to continue moving forwards but with some sideslip velocity.
If we complete a full 360 degree turn with some bank angle the aircraft must experience 360 degrees of rotation about its centre of gravity, so where does this rotation come from?

Welcome to aviation.stackexchange.com!

"I don't understand why this doesn't just cause the aircraft to continue moving forwards but with some sideslip velocity."  It will if you counter the direction of bank with opposite rudder.

The airplane will also rotate around the pitch axis, the more the steeper its bank angle is. By coordinating pitch and yaw rotation, a circular path can be achieved.

Related -- https://aviation.stackexchange.com/q/34271/34686

You are absolutely right that the horizontal force perpendicular to the direction of flight cause a side slip.
The rotation comes from the vertical stabilizer which as soon as a side slip occurs generates a aerodynamic moment which rotates the aircraft into the direction of flight. This ensures that the horizontal component of the lift vector remains perpendicular to the (aerodynamic) direction of flight.

Technically, the side-slip is a secondary effect - the perpendicular force has no vertical component, by definition. The slip is actually caused by the other aero forces being out of balance.

I am not sure I understand your comment @MikeB. The slide slip is caused by the sum of forces (weight and aerodynamic) resulting in a lateral force which changes the direction of the flight path. When the flight path changes direction (turn) but the aircraft heading remains the same, the result is a sideslip.

It might be easier to understand if you try drawing out the vectors, before and after the turn has been initiated. Use arbitrary values, and you will see that after the turn starts, the VERTICAL component reduces - this is where the problem starts.

Key point: in a steady-state turn, net torque on the airplane must be zero.  Net force on the airplane however is not zero-- there must be a horizontal force toward the center of the turn.  This is called a "centripetal" force.  This comes from the banked wing.
This centripetal force plays the same role as does the steady pull of the sun's gravity upon the earth or any other planet-- it causes the moving body's trajectory to curve.
Our intuition from driving cars, steering shopping carts, etc leads us astray because with these vehicles it seems that the net torque acting on the vehicle might not be zero even in a steady-state turn, when we are going around and around in a circle of constant radius, at a constant speed and turn rate.  But that's an illusion that comes from factors like the castor of the wheels, etc.  Net torque is in fact zero in a steady-state turn even in these vehicles, even if we find ourselves having to constantly exert a torque on the steering wheel, push bar of the shopping cart, etc due to these additional complicating factors.
It's true that as we are initiating a turn, or more specifically as the vehicle's rate of rotation about the yaw axis is increasing rather than staying constant, there must be some net torque exerted on the vehicle to overcome the vehicle's yaw rotational inertia.  In an airplane, we can create that torque with the rudder, or we can allow the airplane to sideslip briefly, in which case the force of the sideways airflow against the fin will generate the torque needed to overcome yaw rotational inertia and increase the vehicle's yaw rotation rate as needed to end the sideslip.1
To maintain a steady-state sideslip, which is characterized by a sideways airflow over the aircraft, in most airplanes we need to hold a constant rudder input toward the high wingtip.2  In this case "downwind" yaw torque from the rudder, and the "weathervane" (or "upwind") yaw torque caused by the sideways airflow over the rear fuselage, fixed vertical fin, etc, are essentially "fighting" each other so that the net yaw torque is zero.3  In some sense we can think of the banked wing as "dragging the airplane sideways through the air", but it's important to recognize that the net force on the airplane is actually zero in any linear, constant-speed sideslip-- the horizontal force from the banked wing is opposed by an equal and opposite horizontal force from the sideways airflow colliding with the side of the fuselage and other surfaces of the aircraft.4
Footnotes:

In actual practice, in most airplanes most of the sideslip we see when entering a turn without using the rudder comes from aerodynamic effects (e.g. "adverse yaw"), not yaw rotational inertia.  Yaw rotational inertia may even act to limit the maximum amount of sideslip caused by these aerodynamic effects (e.g. "adverse yaw"), when entering a turn without using a rudder.  But ultimately, if we are not using the rudder to generate a proverse yaw torque-- say for example if we are only using exactly as much rudder is needed to neutralize adverse aerodynamic torques (e.g. "adverse yaw") and no more-- then something must generate some net yaw torque to overcome yaw rotational inertia and start the aircraft rotating about its yaw axis, and this something will most often be the force of the sideways airflow against the vertical fin as the aircraft sideslips to some small degree.  Of course, as an aircraft enters a turn without using the rudder, it will generally not be obvious to the pilot what percent of any sideslip that temporarily develops is due to adverse aerodynamic torques, and what percent is simply due to yaw rotational inertia.

We're focussing on the "big picture" here, and ignoring the slight sideslip  due to aerodynamic damping in the yaw axis that we'll typically see even in a steady-state turn if the rudder is exactly centered.  This can be further compounded, or can be countered, by aerodynamic yaw torques generated by the deflected ailerons, if some aileron deflection is needed to hold the bank angle constant.  So it's only to a first approximation that keeping the rudder exactly centered will result in zero sideslip in a steady-state turn, where yaw rotational inertia is no longer a factor and aerodynamic yaw torques associated with a changing bank angle are also no longer a factor.

Note that only the side area of the fuselage that is aft of the CG contributes to the stabilizing, "weathervane", "upwind" yaw torque during a sideslip.  Side area of the fuselage that is ahead of the CG contributes a yaw torque in the opposite direction, essentially "helping" the rudder keeping the nose yawed to point in a different direction than the airflow is actually coming from, i.e. a different direction that the aircraft is actually travelling through the air.  An interesting manifestation of this: in the Challenger ultralight, which has large removable cockpit doors forward of the CG of the aircraft, a full rudder input will command a significantly larger sideslip angle when the doors are installed than when the doors are absent and the airflow can pass through the cockpit.  Similarly, aircraft on floats, which have a substantial surface area forward of the CG of the aircraft, often have additional tail area added as well in the form of a ventral fin to restore the desired yaw stability characteristics.

It really makes a lot more sense to think of an intentional, linear (non-turning) sideslip as being "caused" by the deflected rudder, not by the banked wing.  The banked wing is just what cancels the sideforce from the airflow hitting the side of the fuselage, so that the net force can be zero and the flight path can be linear rather than curving.  As a demonstration, we could hold full right rudder and slowly vary the bank angle from a right bank to wings-level to a shallow left bank to a steeper left bank, and to a first approximation we'd have the same angle of sideways airflow over the aircraft the whole time.  But the flight path would vary from a tight skidding right turn (caution!), to a slow wings-level right turn (which still can be called a "skidding" turn), to a situation where the left wing is lower but the flight path is slowly curving to the right (still a form of "skid"?), to a linear slip with the left wing low, to a slipping left turn.  From the point of view of the "yaw string", these are all manifestations of sideslip, and the deflected rudder is the fundamental cause of the sustained sideways airflow over the aircraft, not the banked wing.

snall point but Newton's first law applies to rotations and moments in exactly the same way as it applies to forces and velocities. Just as it is true that a body in motion will remain in motion at a constant velocity unless a force is applied, a body in rotation at a constant angular velocity will remain in rotation at the same angular velocity unless a moment is applied to it.

I didn't get into it here in this answer, but the idea that some sustained net force (such as the sideways from the banked wing) could result in some sort of steady-state, linear flight path is basically an Aristotelian concept, rather than a Newtonian one.  Steady-state linear sideslips can only be properly understood when we realize that the sideways force from the banked wing is opposed by another force (as explained in answer), and the net force is actually zero -- .  And once we realize that the sideways airflow over the aircraft is playing a key role in the forces at play, (ctd)

(ctd) then we have to think about how that sideways airflow affects the torques as well. We'll soon realize that since aircraft have "weathercock" or "directional" stability, a linear (non-turning) sideslip generally won't persist for long unless the rudder is deflected. Exceptions exist, where strong dihedral stability requires strong aileron deflection to maintain a constant bank angle while slipping, and the adverse yaw characteristics of the ailerons are such that the aileron deflection itself is sufficient to create a strong, "adverse", "downwind", yaw torque, sustaining the slip. (ctd)

(Ctd) In such an a/c you could actually find yourself in a sustained, constant-banked, steady-state sideslip with the stick or yoke deflected to create a roll torque toward the low wingtip and prevent the bank angle from decreasing, and the rudder centered. I've seen some evidence of such behavior in some some sailplanes.

Although you did not explicitly reference my comment in your comments, if they were in response to mine, let me clarify... The reason I posted it is because I feel that the fundamental aspect of the concept of a "sideslip" is not the change in the direction of aircraft velocity, it is the [lateral] mis-alignment of the aircraft's orientation with it's velocity. Understanding that, is it not clear that sideslip is not caused, (explained), by forces, but by the distribution of forces relative to the CG, i.e., by moments, as you expertly explain in your additional comments.

@CharlesBretana -- actually my comments were not really directed toward yours, just an expansion on my answer

no worries, then, Happy New Year!

You are NOT right. An aircraft in flight has six degrees of freedom. You must analyze them all to understand the motion.
The aicraft is in motion in three dimensions. It can move, in each of these three directions, and it can also rotate in each of those three directions.
To understand the motion, you need to analyze the total forces acting on the aircraft, and resolve them ( add them up vectorially), into a single force representing the sum of all of them. That vector sum of all forces then determines the direction and magnitude of the acceleration the aircraft will experience. if the orientation or direction of that total force is to one side, the aircraft will turn. I don`t mean it will rotate, we'll get to that later. I mean it will turn. It's velocity vector will change direction.
The other three degrees of freedom control what the aircraft attitude is, i.e., how it rotates or what direction it is pointed in.  This is controlled by the torques or moments created by the distribution of the forces on the airframe relative to the center of mass (often called center of gravity or CG) - this is called a torque, or moment. (Sit on one end of a seesaw with your 50 lb granddaughter on the other end)
In the scenario you describe, the horizontal component of the lift vector creates a horizontal component of acceleration, and that causes the velocity vector to change direction. If there was no rudder, the aircraft would still change direction of motion, but it's orientation (the direction the nose is pointed in), might not change. That is controlled by the distribution of the forces on the airframe relative to the CG. If more force is pushing to the right 5 feet ahead of the CG than is pushing right 5 feet behind the CG, then the aircraft will Yaw to the right.
Basically, the moment is the product of the force vector times the distance between the CG and where the force is pushing on the airframe.  You add up all the moments and determine the total moment to see what the impact of all of them on the aircrafts tendency to rotate, (pitch, yaw or roll).
Because stability about the vertical axis (directional stability) is so important, almost all aircraft are designed with aerodynamic surfaces that provide this. We call them vertical stabilizers. And most of them have adjustable sections (rudders) to allow the pilot to control them. This is what affects and controls sideslip, not the horizontal component of lift.

The answers here are correct but let me try to illustrate this in an intuitive way that will “sink in” for most readers even without a physics understanding.
You “get” the part about the bank causing a horizontal component to the wing’s lift, in the direction of the turn.  From that starting point, take the wings and the bank out of it.  Imagine a simple arrow in flight, which is similarly starting to fly “sideways” just like the banked airplane is starting to fly sideways.  What would happen to that arrow?  The tail feathers or “fletching” would cause the arrow to tend to turn into that wind, just like a weather vane.
If you bank the plane left, the aircraft does start to move left, and the tail, just like on the arrow, weather vanes the airplane into the turn.
The technical reason for this is that the CP (center of pressure) is well behind the CG (center of gravity) but hopefully, even if you’ve never heard of either of these concepts, the arrow example helps you form the correct intuition in your mind.

why this doesn't just cause the aircraft to continue to move forward but with some sideslip velocity

we are left with a horizontal component of lift acting perpendicular to the forward direction of the aircraft

Let's start there: 3 force vectors that the aircraft is producing acting against gravity and drag.
We roll into the bank, increasing Angle of Attack (and Power) to produce a larger lift vector to maintain adequate vertical lift.
Mathematically, Lift = 1/cosine bank angle × vertical lift, and the horizontal lift component is sine bank angle × Lift.
The horizontal component and the thrust component act on the plane, the result is a change in course of the aircraft.
Now think of what a weathervane does when the wind changes.  In a relativistic sense (as in wind tunnels) the wind on the stationary weathervane and the plane flying through the air will have the same effect:  the weathervane or the plane will try to align itself to the new "relative wind" direction.

Either the wing, outside aileron, or inadequate rudder (slip) or excessive rudder (skid) can cause the relative wind to be either inside or outside the aircraft direction of flight, respectively.

The rear fuselage of the aircraft and empennage are designed the "weathervane" into the relative wind, and the rudder provides "fine tuning" for "coordinated" flight.
So, the "circular path" is a result of the horizontal lift component --> changes relative wind --> creates sideforce on aircraft --> torques the tail around the Center of Gravity to realign with the relative wind, over and over again.
Notice, as the pilot controlling the plane, you can move the position of the "ball" with either ailerons or rudder.  Usually, we bank a turn to change course, and "clean it up" by positioning the rudder properly to give just the right amount of yaw to coordinate the turn.

keeping the relative wind as close to the nose as possible minimizes drag

So, it seems the trick is to rotate the aircraft around the turn with the rudder instead of using sideforce (from uncoordinated flight) to keep the nose pointed into the relative wind.  An object in orbit can turn without rotating because there is no drag.

Could one imagine trying to "coordinate" the rotation (degrees/second) of the satellite with the orbit time to always have a view of the earth?

Banked flight indeed produces a side force. A side force produces sideslip. Unless it passes through the center of gravity, it will also produce a rotation. Ideally, the total effect is such that a passenger not looking out of the window does not notice anything. Effective gravity should still be directly head to feet. Rudder may be required to sustain the illusion. This may help you to understand some of the responses that you received.indeed

"A side force produces sideslip." -- not always! "Unless it passes through the center of gravity, it will also produce a rotation." -- substitute "torque" for "rotation".  Not the same thing!

Forces do not themselves produce sideslip, if a force is applied to a body so that it passes directly through the body's center of mass, the body will experience an acceleration, and will change its velocity, but will not change its orientation, because no torque or moment will have been generated. It is the torque or moment created by the vertical stabilizer or rudder displacement or p-factor or propellor slipstream that creates sideslip.


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What causes the secondary effect of rudder, and which direction does it act?
I understand that rudder deflection causes several things to happen. Primarily the aircraft yaws, and for a positive (trailing edge left) rudder deflection there will be a positive side force generated and negative yawing moment to yaw the plane nose left. I also understand that there is a direct secondary effect of rudder in that the side force acting at a distance above the roll axis of the aircraft creates a rolling moment (positive rolling moment to starboard for a positive rudder deflection) and there is a yaw-induced effect where as the aircraft yaws (in the case of positive rudder deflection the yaw is to port) the outer (starboard) wing experiences a greater flow velocity and therefore generates more lift which generates a negative rolling moment (to port).
My question is what is the overall rolling moment for some rudder deflection, is the direct effect stronger or is the yaw-induced effect stronger.
For context I am a third year aeronautical engineering student trying to answer this question:
"About which axis does the secondary effect of rudder deflection act? About this axis
which direction (sign) would a moment generated from a negative rudder defection
act, for a conventional tailplane and rudder (relevant control derivative is positive)?
Explain any reasoning behind your answer."

I think you've already answered your question

"acting at a distance above the roll axis of the aircraft creates a rolling moment (positive rolling moment to starboard? for a positive rudder deflection"? If by "positive", you mean right rudder, this would create left roll, not a right roll, because the rudder is mounted above the roll axis, and would create a counter-clockwise roll, (as viewed from behind), just like down aileron on the right wing or up aileron on the left wing.

The total resultant effect on roll from any rudder input, is a complex issue, dependant on several other factors, (AOA, airspeed, wing anhedral/dihedral and sweep, etc.), but generally, it will be in the same direction as rudder input, (right rudder will generate right roll and left rudder, left roll)

It would help a lot to build your own aircraft and try it.  Imagine a very tall thin balsa rudder on your glider, then a square one with equal area.  Then try varying degrees of wing  dihedral.  A small hill and some glue is all you need (paper works too).

@CharlesBretana The z axis positive direction is defined down in the kinematic coordinate system. The signs of deflections, forces and moments are correct.

Normally, rolling will be dominated by dihedral. The rudder will have only a minor influence, but it controls the yaw angle which produces the sideslip which in turn is necessary for dihedral to do its job.

@Peter,  "Normally, rolling will be dominated by dihedral" For straight wing aircraft with dihedral, perhaps. But for swept wing aircraft, perhaps not so much.

@Peter, so then left aileron, causing a left roll, (counter clockwise from behind), is considered a positive rotation? Thank you, I did not know that. Do you happen to know why this convention was adopted? It seems counter-intuitive. In almost all other contexts clockwise rotations are the default.

@CharlesBretana I agree, it is a bit strange. The reason is to have X point forward and Y to the right. To get the coordinate system right, now Z has to point down. Use your right hand: Thumb is X, index finger is Y and middle finger is Z. In other coordinate systems (eg. for the structure) X points back, Y right and now Z can point up.

Ah yes, the ubiquitous Right-Hand-Rule. I always used X to the right, Y up, and Z forward into the paper

If you consider the roll torque generated by the overall shape of the aircraft (including the difference in forward airspeed between the left and right wings) due to the yaw rate induced by the rudder to fall within the scope of your question (a "secondary effect" of the rudder),
then should you not also consider the roll torque generated by the overall shape of the aircraft (especially including dihedral wing geometry, wing sweep, and high vs low wing design) due to the sideslip angle induced by the rudder to also fall within the scope of your question (i.e. a "secondary effect" of the rudder)?
It's not obvious to me how expansive the homework question is trying to be when it mentions the "secondary effect" of rudder.  Perhaps it is only focusing on the direct roll torque created by the rudder, independent of yaw rate and independent of sideslip angle.  In which case the answer is fairly obvious for most aircraft.
But your own specific question here to ASE seems to be whether the "direct" roll torque created by the rudder is typically stronger or weaker than the yaw-rate-related roll torque created by the rudder.  Without directly answering your question, I'd suggest that it may help you understand the overall problem if you realize that the sideslip-related roll torque created by the rudder is typically stronger than either of those other two things.
(It may help you better understand the distinction between yaw-rate-related effects and sideslip-related effects, if you consider the case of a sustained constant-heading sideslip.  To hold the bank angle constant in this maneuver, most aircraft need a fairly significant aileron input "opposite" the rudder input, which tells you that the sideslip-induced roll torque created by the rudder is dominating over the "direct" roll torque created by the rudder.)

In other words-- see Peter Kampf's comment under question ( https://aviation.stackexchange.com/questions/102401/what-causes-the-secondary-effect-of-rudder-and-which-direction-does-it-act#comment286943_102401 )


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Coordinate system in aircraft. Question about step between lines
Good day! Can you help me? I doing a restoration of the plane's geometry. I want to do restoration use a theory. But I have not information about step (size) between B.L. (W.L.) lines. As example - this picture with bulkhead on which are B.L.4.,8.,12...

Given the general size of what you're restoring, is it more reasonable that the dimensions are shown in inches or in centimeters? The overall height seems to be about 68 units (i.e. 27.47 of them above the waterline, and 40.53 of them below the waterline), and the max width about 53 units (i.e. 26.66 of them to either side of the centerline). The lines themselves seem to be 4 units apart, both laterally and vertically. So if you can tell us if this is all in inches or centimeters (or something else, possibly) we can discuss everything else with specifics.

@RalphJ. In this blueprint are inches

The Vertical Centerline, BL4, BL8, BL12, and BL16 lines are each 4 inches apart; the WL20, WL16, WL12, and so on down to the Horizontal Centerline, aka the Waterline, are also all 4 inches apart.
So if you picture an X-Y axis with the origin of 0,0 at the point where the Vertical Centerline and the Horizontal Centerline meet...

The top center point would be at 0,27.47" (0 because it's on centerline, and 27.47 from the dimension showing the max height -- to the left of the drawing)
The next point on the outline, working clockwise, would be at 4, 27.159 (4 because the BL4 is 4 inches from the vertical centerline, and 27.159 because that's what's shown under BL4)
The next points would be 8,26.204 then 12, 24.530 then 16,21.973

After that, the points are no longer every 4 inches horizontally, but every 4 inches vertically:

18.275, 20 (using the distance from the Vertical CL on the WL20 line)
then 21.671, 16 then 23.982,12

and so on until reaching the Horizontal CL

26.660, 0 is the point on the horizontal CL

After that, the X values start to decrease, and the Y values become increasingly negative:

26.530,-4
26.136, -8

And so on down to the last WL line at -32 inches:

16.362, -32

Now the lines go back to every 4 inches horizontally:

16, -32.42 (very close to the prior point)
12, -36.192
8, -38.662
4, -40.071

And finally, the point at the bottom on centerline:

0, -40.530

And those are the points that the diagram provides for you.


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Vertical gaps between stairs and aircraft entrance
I have gotten an answer saying that the stairs are not an universal height and can be adjusted due to passenger weight and the weight on the plane. What if this gap occurs as the passengers are climbing into the aircraft? As I would expect that the plane would be heavier and therefore a gap where the airstairs are below the entrance is unusual?
Additionally, what is the mechanism that is used to adjust the stairs to fit to the height of the plane?

This seems like a response to something. What’s the missing context?

I’m voting to close this question because it's a requesting clarity for an answer to a previous question, therefore should be a comment to that answer rather than posed as a new question.

See this:  https://aviation.stackexchange.com/questions/102923/if-the-stairs-are-a-universal-height-how-does-a-vertical-gap-occur-e-g-where

Stairs are of course not of fixed height and they can be adjusted either mechanically or hydraulically. Though weight of empty aircraft and that of fully laden aircraft may greatly vary, the height of cabin entrance changes by only few centimeters. (only as much as the landing gears and tyres can get compressed). As the passengers get in, this depression in height is not even observable. The stairs are positioned at the cabin entrance at a pre-determined height specified for the type of aircraft taking into account the possible reduction in height of the cabin entrance. So it really does not matter, how many passengers get in or how much is the laden weight of the aircraft. The difference in height of cabin entrance and the top rung of the stairs will always be much less than a few centimeters and average person can easily negotiate without even acknowledging the difference.


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Question about the gap between the airstairs and the aircraft
Is there universal code/height between the ramp at the top of the airstairs and the aircraft. If not, what is the expected gap between the aircraft and the airstairs when embarking or disembarking?

Are you asking about the horizontal gap or about the height difference?

Sometimes, there is a gap between the air stairs ramp and the aircraft. It can be due to some reasons like

Preventing damage to the fuselage
The ramp not fitting all the
aircraft as their fuselage dimensions are different.

According to FAA Advisory Circular 150/5220-21C,

No gap between vehicle and ramp will exceed 5/8 in (16 mm).

Related question: What is the maximum acceptable gap between the jet bridge and airplane door?


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If the stairs are a universal height, how does a vertical gap occur? E.g. where the stairs fall short of the aircraft fuselage
If the stairs are a universal height, how does a vertical gap occur? E.g. where the stairs fall short of the aircraft fuselage.
I have tried to find information regarding whether the stairs are universal in height and whether they can fall short leaving a vertical gap between the stairs and the aircraft.

Do you ask about the stairs at the Airport, that drive to the aircraft? Your question is a little unspecific in that regard.

Aircraft aren't exactly "at a universal height." Even for a given type, as passengers deplane, the struts extend, and the door can rise several inches going from full to empty. Is this perhaps the dynamic you're observing?

Maybe a duplicate of https://aviation.stackexchange.com/questions/101290/height-from-the-ground-of-aircrafts-doors

Aircraft stairs are not a universal height. They almost all have some sort of mechanism that allows them to be adjusted to be used with different size aircraft.
The reason a vertical gap between the top of the stairs and the aircraft entrance occurs is because the aircraft suspension (for lack of a better word) will decompress as passengers deboard and the plane becomes lighter, which causes the aircraft body to rise. The opposite happens as passengers board and the plane becomes heavier.

while passenger weight is a factor, so is fuel and cargo. The struts on the landing gear will compress with additional weight. Stairs and jetways do have the ability to adjust their height. The person operating the jetway or stairs often get the height close enough and does not adjust it exactly.

Thank you for your answers. What if this gap occurs as the passengers are climbing into the aircraft? As I would expect that the plane would be heavier and therefore a gap where the airstairs are below the entrance is unusual?

What if? If it becomes excessive the crew should adjust it, otherwise passengers simply move their feet as needed.  I'm not sure what you are confused about here...


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Where can I ask questions (career advice) about jobs in Aviation Management?
I am thinking of changing my career. I have a physics degree and I want to do a masters in aviation management.Is there any sites where can I get an idea about my chances at getting a job in the aviation management sector, if I do a masters? (with no prior experience).

I'm voting to close this question as off-topic because this question is about locating resources for career advice.

@kevin I can see your point, but I kind of disagree.  We do cover questions about the operational side of airlines, and getting a job with operations would fit under that (I think).  I'm voting to reopen it, but I'll also open a question on meta to see what everyone thinks.

The discussion is here: http://meta.aviation.stackexchange.com/questions/3067/are-questions-about-careers-in-aviation-on-topic

There is a huge demand for aviation professionals. However, how that demand is looks like is very difficult to answer, since it's a pretty fragmented market and getting an overview is very difficult.
As such, you're best chances of understanding it are:

Going to recruitment fairs that major airlines attend.
Attending events that airlines hold themselves.
Trying to get in touch with human resources in an appropriate way directly.

From this you should be to gauge how interesting a candidate you will be. As with most things, this is hardly an exact science. Getting a program with a placement/internship as part of the course for instance might be quite beneficial. As for the no previous experience, I think this is the case for most people heading into this area.

Thank you for your answer! I was also wondering, if you know any website especially catered to aviation careers? Maybe I can ask my question there?

I think you're unlikely to find a forum for this sort of thing. I don't know where you are geographically, but try something along for instance Lufthansa Careers or their facebook page be-lufthansa. Should be plenty of others as well. If it's fairly exact/well defined, you can try asking it here as well as a new question.

thank you! this site was indeed helpful! thank you again for taking the time to reply me :) have a great day ahead!

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


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What types of aircraft can TS-1 fuel be used for?
I'm curious to learn about what types of military aircraft can take TS-1 fuel. For example, can a SU-30 or a YAK-130 use TS-1, or do they only take A-1? Also what types of helicopters can take TS-1?
Can TS-1 be used in any civilian aircraft?

TS-1 can be used in any subsonic turbine-powered aircraft.
Note that the second of your questions has already been asked: Can TS-1 fuel be used in an aircraft that takes Jet A?
I'll add a brief note - all civilian jet aircraft are subsonic, so they can technically operate on TS-1 fuel, even if only approved for RT. Unapproved but compatible fuels can result in shortened maintenance intervals.
Here is a brief list of all Russian jet fuel types. TS-1 has a very low freezing point, below -60C, allowing for cold weather operation. But the lower the freezing point, the lower the flashpoint, so a light fuel like TS-1 is more hazardous than Jet-A in hot weather. RT fuel is used for that.
For the same reason, TS-1 is not recommended for supersonic flight, where the fuel heats up and light fuels can boil off. If combat is possible on the mission, the Su-30 and other supersonic aircraft need to use the more versatile RT fuel, which has a higher boiling temperature.
Supercruising aircraft like the Tu-160 or SR-71 get even hotter, so they need heavier fuels, such as T-6, T-8 or JP-7, that can take higher temperatures. These fuels are not suitable for subsonic civilian aircraft.


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The F-16 has a FLCS that constantly holds 1 g. What happens on high pitch angles?
The F-16 has a FLCS that constantly holds 1 g. What happens on high pitch angles?
What is defined as 1 g? let us say that you climb with 60 degrees relative to earth, does that mean that the FLCS just registers 0.5g and trims the plane up?
What happens at 90 degree pitch? Does it trim a whole G?
This is what DCS and BMS currently do and it doesn't make sense to me.
(This is all with AOA of less than 10 degrees)

Ask why an aircraft needs an angle of attack:  to make lift against gravity.  The four forces must be in balance to fly a straight line.  In a steep climb AoA is reduced because gravity vector (WRT the aircraft) is now more aligned with drag, requiring the engine to do more work.  In a 90 degree climb "lift" will only curve the flight path.  In order to fly straight, the computer sets AoA to zero lift and the engine does all the work, essentially as an airbreathing rocket.

The gravity by definition always exerts 1 g of accerleration on any given object. If this acceleration is compensated by a counterforce, the object is weightless. Therefore if no other forces are acting on the object, the object is unaccelerated and continues on its current trajectory in a straight line.
Consider the example of the F-16 in straight an level flight and plot the different forces you arrive at the following well-known picture:

The complete gravity of 1 g is compensated by the lift of the aircraft and drag is compensated by thrust.
During a climb the situation is a bit different:

As you can see, the gravity still pulls the plane down towards earth, but now the lift and thrust combined cancel out the gravity combined with the drag of the aircraft. However the magnitude of the gravity is still 1 g.
But the FLCS of the F-16 does not do this for you as it always commands 1 g of normal acceleration (acceleration vertical to the body of the aircraft, $n_z$) regardless of pitch attitude (according to this source). Therefore during a climb you would have to trim (reduce) the normal acceleration $n_z$ commanded by the FLCS such that a stable climb is achieved. For example in a $60^\circ$ climb, you would have to reduce the normal acceleration command to only $\cos(60^\circ)\cdot 1 g = 0.5g$ and adjust the throttle such that more thrust is generated. This is not automatically done by FLCS!
However in newer blocks, the FLCS automatically compensated for roll attitude. Therefore if you fly a curve with a bank angle of say $60^\circ$, the acceleration command is automatically increased to 2 gs to compensate for the centrifugal acceleration of said curve. However apparently this is only implemented in newer F-16 models.
With this in mind, the target acceleration formula for hands-off stick flight, looks something like this:
$n_{z,cmd} = 1g\cdot sin(\varphi)\cdot \delta_{trim}$
with $n_{z,cmd}$ being the target normal acceleration (acceleration vertical to the body of the aircraft), $\varphi$ the roll angle, and $\delta_{trim}$ the trim input, normally being 1.

Thanks for the answer. So yeah, it makes sense to me too.
But your sources isn't really helping, it seems to be 2 comments from a forum where both are unsure.
At high pitch angles, the accelerometers would need to compensate for thrust and drag, making the calculation slightly more complicated. Is that something the 50 year old F-16 could do?

@Sandpatch I included the link to the forum because both of them are sure that the pitch attitude is not compensated. Therefore the FLCS will command 1 g regardless of pitch attitude. Again, I cannot give you a flight handbook as a source, but consider this video. Regarding the computational power: The flight law they are proposing is a simple cosine calculation. The FLCS has to perform a lot more calculation than this to achieve stable flight. Therefore, no this would not be a problem.

@Sandpatch but to reitterate: The FLCS always commands 1 g of upwards acceleration (therefore upwards in regards to the pilot, regardless of orientation of the aircraft) unless the pilot trims otherwise! If you want the aircraft to climb say at 60°, the pilot has to manually trim the aircraft such that the target acceleration is only $cos(60^\circ)\cdot 1 g=0.5 g$ and adjust the throttle accordingly. Perhaps I should include this in my answer to make this perfectly clear...

@Sandpatch updated my answer accordingly

@U-flow Thanks for the answer. Seems like it is like that.

Re "Therefore during a climb you would have to trim (reduce) the acceleration commanded by the FLCS such that a stable climb is achieved. For example in a 60∘
climb, you would have to reduce the acceleration command to only cos(60∘)⋅1g=0.5g
and adjust the throttle such that more thrust is generated!" -- you seem to be using "acceleration command" to mean "the lift force commanded" (divided by mass).  Since thrust and drag can also exert force components which can be correlated to acceleration components, the answer is a little confusing.  Not sure what to suggest to make it clearer.

Maybe saying "the Nz acceleration commanded" would be clearer (see recent other answer for context)

I tried to add this context by editing accordingly

In the two images you might want to take AOA into account, as the Lift vector is perpendicular to the velocity vector in flight dynamics. The images are only true if AOA=0.

You are absolutely right. But for the sake of intuitive simplicity, I will leave it as is

The FLCS holds 1 g in absence of stick input.
This 1 g is maintained in relation to Earth. The result is that, given enough thrust and lift, the airplane's nose will keep pointing where the pilot has last pointed it. This is maintained whether pitch is up or down, or when the plane is in a coordinated turn.
When climbing at a 60 degrees up pitch, the airplane is still at ~1 g, which simply means it's not rapidly accelerating in any direction. Climb isn't acceleration per se. Whether at 15 degrees or 90 degrees, it just requires the forces to balance out while maintaining some upward velocity component.
Here's a detailed explanation of climb physics: Does lift equal weight in a climb?

"When climbing at a 60 degrees up pitch, the airplane is still at ~1 g,"
It's not, or at least it depends on your definition of G. At 60 deg pitch up, in steady state, the IRUs would measure 0.5 vertical G, and sqrt(2)/2 longitudinal G. If measured in earth axis, the up/down component would indeed be 1. OP is asking if the FLCS feedback is based upon body-axis G or earth axis G.

@sbabbi It's earth axis G.

Thanks for the answer, but this was a question specific for the F-16 FLCS. Does the F16 also take this into account because it would have to subtract acceleration and braking forces to the accelerometers reading, making it slightly more complicated.

@Therac From what I can gather from online sources, the commanded g-rate is independent from the pitch attitude, and only (in newer blocks) is dependent on the roll attitude. Consider this video from DCS or this source. This means that for example during a climb, the pilot would have to trim the aircraft to command less then 1 g. If I am not mistaken, your answer implies that this is done automatically, however this is not the case. For example in inverted attitude the plane pulls down with now 2 g!

@U_flow You command any g you need with the stick, and while you're doing it, it's anything within the envelope. Once released, the plane stabilizes at 1 g in relation to Earth. At least that's the idea, implementations aren't all perfect. But DCS seems to do just that.

No F-16s ever uses earth axis G as feedback, nor F-18 or F-35. According to the DFLCS control block diagram from the F-16 System Program Office (see the answer below), the F-16 is specifically using Nz (normal load factor) feedback. This is the body axis G by definition. Imagine the aircraft being inverted, it should be pulled towards the ground by the FLCS.

It seems that I'm late to the party, but the key reason is that the F-16, F-18, F-35, etc. uses Normal Acceleration, or Normal Load Factor feedback in the cruise or Up & Away (UA) flight mode. It does not compensate for pitch or roll. (The F-18 and F-35 however, will shift from Nz command to pitch-rate command at lower speeds.)
Definition: The Normal Acceleration (an) or Normal Load Factor (Nz) is the acceleration in “g” along the aircraft body Z axis, not including the components of gravity.
If you're climbing vertically with a 90 deg pitch, your Nz will be 0g. So the Flight Control System will constantly pull the aircraft to maintain Nz=1g.
Just be aware of in which axis system the acceleration you're talking about.

The ultimate source of F-16 FLCS using Nz feedback:

(F-16 Digital Flight Control System Functional Block Diagrams. Data obtained from Lt. Bruce Peet, F-16 System Program Office, Wright-Patterson OH.) Quoted in the paper DTIC ADA 189675.
Document Link: https://apps.dtic.mil/sti/pdfs/ADA189675.pdf

Re "including the components of gravity" -- did you maybe mean not including the component of gravity?  If gravity is included, then you have to say the situation in straight and level flight for the Nz component is 0G not 1G.

@quiet flyer Yep I missed that. Thanks.

+1 for the control scheme. Could you perhaps also add the link to the source?

@U_flow Sure, i've included it in the answer.